【BZOJ2223】【SPOJ】PATULJCI(主席树)

Description


Solution

主席树板子。
对于每个区间找左右子树哪一个大于 r l + 1 2 ,递归求解即可。


Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ2333 & BZOJ3524] PATULJCI
 * Date: Apr 13th, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 300005, maxc = 100005;

int rt[maxn], d, cnt, Ans;

struct Node
{
    int l, r, val;
}tr[maxn * 33];

#define mid ((l + r) >> 1)

int update(int las, int l, int r, int x)
{
    register int now = ++ cnt;
    tr[now] = tr[las], ++ tr[now].val;
    if (l != r)
        if (x <= mid) tr[now].l = update(tr[las].l, l, mid, x);
        else tr[now].r = update(tr[las].r, mid + 1, r, x);
    return now;
}

int query(int lrt, int rrt, int l, int r)
{
    if (l == r) return Ans = l;
    if (tr[tr[rrt].l].val - tr[tr[lrt].l].val > d) return query(tr[lrt].l, tr[rrt].l, l, mid);
    if (tr[tr[rrt].r].val - tr[tr[lrt].r].val > d) return query(tr[lrt].r, tr[rrt].r, mid + 1, r);
    return 0;
}

int main()
{
#ifdef hany01
    File("bzoj2223");
#endif

    static int n = read(), c = read(), tmp;
    For(i, 1, n)
        rt[i] = update(rt[i - 1], 1, c, tmp = read());

    for (static int m = read(), x, y; m --; )
    {
        x = read(), y = read(), d = (y - x + 1) >> 1;
        if (query(rt[x - 1], rt[y], 1, c)) printf("yes %d\n", Ans);
        else puts("no");
    }

    return 0;
}

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转载自blog.csdn.net/hhaannyyii/article/details/79932963